How do I find real zeros?

Question

Solve in the real number system:

6x^3 + 11x^2 - 4x - 4 = 0

How? Thank you.

Answer 1

This is the approach to factorising and solving cubic equations.
1) Using trail and error, determine the first factor of the polynomial.
2) Using synthetic division, determine the remaining factor in the form of a quadratic equation.
3) With a quadratic equation as a remaining factor, factorise the equation if possible or apply quadratic formula to find the roots of the equation.

Let f(x) = 6x^3 + 11x^2 - 4x - 4

Using trail and error, the first factor of f(x) is (x + 2).

By synthetic division,
-2.|..6..11..-4..-4.
....|.....-12...2...4.
....---------------------
......6..-1...-2....0.

Rewriting the polynomial f(x),
f(x) = (x + 2)(6x^2 - x - 2)
f(x) = (x + 2)(3x - 2)(2x + 1)

when f(x) = 0,
(x + 2)(3x - 2)(2x + 1) = 0
x = -2 or x = 2/3 or x = -1/2

Therefore, the roots to f(x) are -2, 2/3 and -1/2.

NOTE: The details of synthetic division are on
http://www.purplemath.com/modules/synthdiv.htm

Answer 2

Solve,

6x^3 + 11x^2 - 4x - 4 = 0

Test x = -2

6(-2)^3 + 11(-2)^2 - 4(-2) - 4

= 6(-8) + 11(4) + 8 - 4

= - 48 + 44 + 8 - 4

= - 52 + 52 = 0

So, x = -2 is a solution,
hence (x + 2) is a factor.

use long division (6x^3 + 11x^2 - 4x - 4) divided by (x + 2), to get the factors:

(x + 2)(6x^2 - x - 2) = 0

(x + 2)(3x - 2)(2x + 1) = 0

So x = -2, x = 2/3 and x = -1/2

Answer 3

The possible roots are +/-1,+/-2,+/-4,+/-1/2,+/-1/3,+/-1/6,+/-2/3,+/-4/3...
by trial and error, and by synthetic division..
6 11 -4 -4 | 2/3
4 10 4
6 15 6 0

the constants at the bottom, not including zero (0) are the coefficients of the depressed equation. That would be
6x^2 + 15x + 6 = 0
dividing all the terms by 3, we have
2x^2 + 5x + 2 = 0
(2x + 1)(x + 2) = 0
2x + 1 = 0 ; x + 2 = 0
x = -1/2 ; x = -2

Thus, the roots are 2/3, -1/2 and -2

Answer 4

6x^3 + 11x^2 - 4x - 4 = 0
6x^3 + 12x^2 - x^2 -4x - 4 = 0
6x^2(x + 2) - (x + 2)(x + 2) = 0
(6x^2 - x - 2)(x + 2) = 0
(6x^2 - 4x + 3x -2)(x +2) = 0
(2x(3x -2) + (3x -2))(x + 2) = 0
(2x +1)(3x -2)(x +2) = 0
x = -1/2
x = 2/3
x = -2

Answer 5

? polynomial f1(x) is of surprising skill; polynomials of surprising skill have uncomplicated variety of real roots; df1(x)/dx = 21x^2 +5 > 0, for this reason end: max/min aspects do no longer exist, for this reason f1(x) has 2 complicated roots and one real root; for that reason x1=a, x2=b-j*c, x3=b+c*j; in accordance to Viet theorem we get: x1*x2*x3 = minus coefficient at skill 0 this could nicely be a*(b-jc)*(b+j*c)=-(-a million); or; a*(b^2+c^2) =a million, for this reason a > 0; the real root is beneficial! ? polynomial f2(x) is of even skill; polynomials of even skill have 0 or perchance variety of real roots; df2/dx= -16x^3 -2x =0; for this reason we get basically one extreme factor x1=0, and this factor is for optimal of f2; f2(x1)=-4*0^4 –0^2 –0 –a million = -a million < 0; this implies the finished f2(x)<0, for this reason f2 has no real roots, all 4 roots are complicated roots!

Answer 6

If the equation is not factorable, graphing is the easiest way; try synthetic division next.

Answer 7

by trial and error method.
for x=-2 LHS=0
Hence -2 is a root.
Use horner method to find the other roots